mathop {lim }limits_{h to 0} frac{{2left[ {sq | vedclass

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(D) We have the limit: $\mathop {\lim }\limits_{h 	o 0} \frac{{2\left[ {\sqrt 3 \sin \left( {\frac{\pi }{6} + h} ight) - \cos \left( {\frac{\pi }{6

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$\frac{4}{3}$

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(D) We have the limit: $\mathop {\lim }\limits_{h 	o 0} \frac{{2\left[ {\sqrt 3 \sin \left( {\frac{\pi }{6} + h} ight) - \cos \left( {\frac{\pi }{6} + h} ight)} ight]}}{{\sqrt 3 h(\sqrt 3 \cos h - \sin h)}}$ Multiply the numerator and denominator by $\frac{1}{2}$ inside the bracket: $= \mathop {\lim }\limits_{h 	o 0} \frac{{4\left[ {\frac{\sqrt 3}{2} \sin \left( {\frac{\pi }{6} + h} ight) - \frac{1}{2} \cos \left( {\frac{\pi }{6} + h} ight)} ight]}}{{\sqrt 3 h(\sqrt 3 \cos h - \sin h)}}$ Using $\sin(A-B) = \sin A \cos B - \cos A \sin B$,where $A = \frac{\pi}{6} + h$ and $B = \frac{\pi}{6}$: $= \mathop {\lim }\limits_{h 	o 0} \frac{{4 \sin \left( {\frac{\pi }{6} + h - \frac{\pi }{6}} ight)}}{{\sqrt 3 h(\sqrt 3 \cos h - \sin h)}}$ $= \mathop {\lim }\limits_{h 	o 0} \frac{{4 \sin h}}{{\sqrt 3 h(\sqrt 3 \cos h - \sin h)}}$ $= \frac{4}{\sqrt 3} 	imes \left( \mathop {\lim }\limits_{h 	o 0} \frac{\sin h}{h} ight) 	imes \left( \mathop {\lim }\limits_{h 	o 0} \frac{1}{{\sqrt 3 \cos h - \sin h}} ight)$ $= \frac{4}{\sqrt 3} 	imes 1 	imes \frac{1}{{\sqrt 3(1) - 0}} = \frac{4}{\sqrt 3} 	imes \frac{1}{\sqrt 3} = \frac{4}{3}$.

$\mathop {\lim }\limits_{h \to 0} \frac{{2\left[ {\sqrt 3 \sin \left( {\frac{\pi }{6} + h} \right) - \cos \left( {\frac{\pi }{6} + h} \right)} \right]}}{{\sqrt 3 h(\sqrt 3 \cos h - \sin h)}} = $

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